1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Electrical Circuit Question

  1. Jan 30, 2013 #1
    http://img15.imageshack.us/img15/771/captureuce.png [Broken]
    1. The problem statement, all variables and given/known data

    Determine if the circuit is valid or invalid.


    2. Relevant equations



    3. The attempt at a solution

    Please note which node I used as the ground node and which node I have labeled [itex]V_{1}[/itex]. By the node voltage method

    [itex]\frac{V_{1}}{8} + 10 - 2 + \frac{V_{1} - 10}{5} = 0[/itex]
    [itex]\frac{13V_{1}}{40} + 6 = 0 [/itex]
    [itex]\frac{13V_{1}}{40} = -6[/itex]
    [itex]V_{1} = -\frac{240}{13} V[/itex]

    [itex]I_{1} = \frac{V_{1}}{8} = -\frac{240}{13*8} = -\frac{30}{13} A[/itex]
    [itex]I_{2} = \frac{V_{1}-10}{5} = \frac{\frac{-240}{13}-10}{5} = -\frac{74}{13}[/itex]
    [itex]I_{3} = 10 + I_{2} = 10 + {-\frac{74}{13}} = \frac{56}{13}[/itex]

    [itex]V = IR[/itex]
    [itex]P = IV[/itex]
    [itex]P = I^{2}R[/itex]

    [itex]P_{8Ω} = I^{2}R = (-\frac{30}{13})^{2}8 = \frac{7200}{169}[/itex]
    [itex]P_{5Ω} = I^{2}R = (-\frac{74}{13})^{2}5 = \frac{27380}{169}[/itex]
    [itex]P_{2Ω} = I^{2}R = (\frac{56}{13})^{2}2 = \frac{6272}{169}[/itex]
    [itex]P_{10V} = IV = \frac{56}{13}10 = \frac{560}{13}[/itex]
    [itex]P_{2A} = \frac{240}{13}2 = IV = \frac{480}{13}[/itex]
    [itex]P_{10A} = IV = 10{-\frac{240}{13}-\frac{74}{13}} = \frac{3140}{13}[/itex]

    These don't add up to zero. I'm kind of confused as to when you are supposed to use a negative sign or not. I used a negative sign in the last calculation because the current is going from lower potential to higher potential. Thanks for nay help you can provide.
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Jan 30, 2013 #2

    gneill

    User Avatar

    Staff: Mentor

    In your first equation, the final term ##\frac{V_1 - 10}{5}## is not correct. The point where the 5Ω resistor terminates is not at 10V above reference since there's a series resistor of 2Ω in the path. You have two nodes to deal with.
     
    Last edited: Jan 30, 2013
  4. Jan 30, 2013 #3
    Hi gneill,

    Why can't I count the node where the 10 volt source, 5 ohm resistor, and 10 Amp source meet to be 10 volts?
     
  5. Jan 30, 2013 #4

    gneill

    User Avatar

    Staff: Mentor

    You could if the reference node was at the bottom of the 10V source. But it isn't. There's a 2Ω resistor in the way.
     
  6. Jan 30, 2013 #5
    Ok thanks I forgot about that resistor.
     
  7. Jan 30, 2013 #6
    http://img59.imageshack.us/img59/1566/capturekwm.png [Broken]

    I believe the picture in this post is accurate. However how would I calculated the power dissipated in the 2A source?

    I know I use

    P = IV
    P = 2(0-(-19.2))
    or
    P = 2(-19.2-0)

    which one is correct?

    Thanks
     
    Last edited by a moderator: May 6, 2017
  8. Jan 30, 2013 #7

    gneill

    User Avatar

    Staff: Mentor

    Your node voltages and currents don't look right. Better recheck your calculations.

    If the potential across a current source is negative (if it's driving a current from a higher potential to a lower potential), then it's absorbing energy.
     
  9. Jan 31, 2013 #8
    What exactly dosen't look right?
     
  10. Jan 31, 2013 #9

    gneill

    User Avatar

    Staff: Mentor

    All the voltages and currents that you've added to the drawing.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Electrical Circuit Question
Loading...